Hypothesis Testing - Chi Squared Test

The Chi-Square test statistic is useful for measuring how close counts of categorical variables are to what we would expect under some assumption (which we will call the Null Hypothesis).

accept or reject the null hypothesis ..

Chi Square Test | P Value | Statistical Hypothesis Testing

In this case, the observed numbers were so different from the expected 1:1 ratio that Yates correction made little difference - it only reduced the X2 value from 13.34 to 12.67. But there would be other cases where Yates correction would make the difference between acceptance or rejection of the null hypothesis.

Null hypothesis for a chi-square goodness of fit test

The resulting F1 offspring from this OH88119 x 6.8068 all showed resistance to bacterial spot. These were then allowed to self-pollinate and a hypothesis was devised to describe the mode of inheritance for the Rx-4 gene. The breeders’ hypothesis was that Rx-4 expression is due to a single gene with complete dominance. The results in the F2 generation though, deviated from what the researchers in that there were more susceptible F2 plants than they anticipated. The question remains, as to how far can observed results deviate from what was expected before they should be considered significant? In other words, is the difference in the number of susceptible plants seen and those expected simply due to chance or is something else going on such as a wrong genetic hypothesis, high influence of the environment, etc.? An example of differences due to chance in this tomato example might be a plant that didn’t get inoculated properly, a worker accidentally hoed out a plant, an incorrectly categorized phenotypic observation etc. So these would be random events which impact results, but are unrelated to the Rx-4 gene. This is where statistics, and in this case study, chi-square comes into play - to help plant breeders accurately interpret results.

Null hypothesis for a chi-square goodness of fit test ..

As noted previously, the general rule in the behavioral sciences is to reject the null hypothesis if the probability associated with the appropriate statistic is less than .05. In this example, p is less than .001 (which is considerably less than .05). Therefore, we will reject the null hypothesis and accept our research hypothesis that there is a difference in term preference between the two groups.

Writing null hypothesis for chi square - I Help to Study

We are now ready for the final step, interpreting the results of our chi-square calculation. For this we will need to consult a Chi-Square Distribution Table. This is a probability table of selected values of X2 (Table 3). Statisticians calculate certain possibilities of occurrence (P values) for a X2 value depending on . Degrees of freedom is simply the number of classes that can vary independently minus one, (n-1). In this case the degrees of freedom = 1 because we have 2 phenotype classes: resistant and susceptible.

The calculated value of X2 from our results can be compared to the values in the table aligned with the specific degrees of freedom we have. This will tell us the probability that the deviations (between what we to see and what we actually saw) are due to chance alone and our hypothesis or model can be supported. In our example, the X2 value of 1.2335 and degrees of freedom of 1 are associated with a P value of less than 0.50, but greater than 0.25 (Follow blue dotted line and arrows in Fig 5). This means that a value this large or larger (or differences between expected and numbers this great or greater) would occur simply by chance between 25% and 50% of the time. By convention biologists often use the 5.0% value (p(See red circle on Fig 5.) If your chi-square calculated value is greater than the chi-square critical value, then you reject your null hypothesis. If your chi-square calculated value is less than the chi-square critical value, then you "fail to reject" your null hypothesis. Therefore in our tomato breeding example, we failed to reject our hypothesis that resistance to bacterial spot in this set of crosses is due to a single dominantly inherited gene (Rx-4). We can assume that the deviations we saw between what we expected and actually observed in terms of the number of resistant and susceptible plants could be due to mere chance. We can continue working with our current hypothesis. Remember, we have not “proven” our hypothesis at this point. Further testing in other crosses and populations will be used to provide additional evidence that our hypothesis accurately explains the mode of inheritance for Rx-4. Click on image below to launch a video tutorial showing how to use a chi-square distribution chart, using this tomato breeding example. Next we will go through a genotyping example, or you can to skip to a discussion about computer programs available when you have extensive data, as well as the strengths and weaknesses of the chi-square test.

Statistics: Chi-square Test of Independence, Null Hypothesis

In the χ2 goodness-of-fit test, we conclude that either the distribution specified in H0 is false (when we reject H0) or that we do not have sufficient evidence to show that the distribution specified in H0 is false (when we fail to reject H0). Here, we reject H0 and concluded that the distribution of responses to the exercise question following the implementation of the health promotion campaign was not the same as the distribution prior. The test itself does not provide details of how the distribution has shifted. A comparison of the observed and expected frequencies will provide some insight into the shift (when the null hypothesis is rejected). Does it appear that the health promotion campaign was effective?